I'm reading An Introduction to Stochastic Processes in Physics by Don S Lemons. Problem 10.2 leads to a pair of equations:

$dV_x = -\gamma V_xdt+V_y\Omega dt-V_y\sqrt{2\gamma dt}N_t(0,1)$

$dV_y = -\gamma V_ydt-V_x\Omega dt+V_x\sqrt{2\gamma dt}N_t(0,1)$

($\gamma$, $\Omega$ constants, $X$, $Y$ position, $V_x$, $V_y$ velocity. The two noises are identical.)

It then asks for the application of the Smoluchowski approximation to these equations. As the book is very hand-wavey in justifying the use of this approximation I find it tricky to reason with. I only know how to use the search-and-replace rules the book gives, eg. set $dV_x=0=dV_y$ and replace $V_xdt$ and $V_ydt$ with $dX$ and $dY$. So what do I do with the terms like $V_y\sqrt{2\gamma dt}N_t(0,1)$ for the Smoluchowksi approximation? Presumably all references to $V_x$ and $V_y$ need to be eliminated.

  • $\begingroup$ Disregarding the mathematical question momentarily: if this is a physics problem, the V factors in front of the noise are very strange. The result is scale invariant in V, so that there is no natural velocity scale, and this is crazy for a thermal system. The resulting brownian motion slowly peters out and stops at long times. It would help if you said exactly what each symbol was--- for those of us that don't have the book's convention handy. I figure $\gamma$ is the damping $\omega$ is the cyclotron frequency, and $N_t$ is white noise. $\endgroup$ – Ron Maimon Aug 21 '11 at 14:20
  • $\begingroup$ @Ron: this is absolutely standard stochastic PDE, see en.wikipedia.org/wiki/Geometric_Brownian_motion $\endgroup$ – Marek Aug 21 '11 at 18:14
  • $\begingroup$ @Marek: It may be standard in finance, but these types of velocity multiplicative noise do not occur in physical Brownian motion, and the title of the question is "cross field diffusion", which I guessed meant diffusion in a B field. The Brownian motion has to settle to a Maxwell Boltzmann velocity distribution. I found the solution to a variant (where the vx and vy noise are identical) yesterday, and it slows down and stops. The correct physics is given by the same equation without the V factors in front of the noise. If this comes from finance, the natural approximations are very different. $\endgroup$ – Ron Maimon Aug 21 '11 at 19:21
  • $\begingroup$ Also, the Smoluchowski approximation is reducing the second order stochastic equation to first order, which is valid at long times compared to how long the velocity distribution takes to relax to Maxwell Boltzmann. This doesn't relax. Perhaps he means omega huge, gamma small perhaps gamma/omega are teeny-tiny. I can't tell which limit is relevant. $\endgroup$ – Ron Maimon Aug 21 '11 at 20:22
  • $\begingroup$ @Ron Yes, the two noise functions are identical, I should have said that. (The book uses different subscripts to indicate independent noise.) What do you mean when you say the physics is that same without the V term. You can't simply erase the V terms as it would have the wrong dimensions. Maybe you can write a full answer. $\endgroup$ – Dan Piponi Aug 22 '11 at 22:52

Original wrong stuff

Well, the OP asked for a solution, but I still can't figure out how this is physics. Add the two equations with an i to get the equation for $V=V_x + i V_y$ (dividing by dt--- I hate finance/mathematics conventions, and calling the white noise $\eta$)

${dV\over dt} = (-\gamma+i\Omega)V + iV\eta$

So that the complex logarithm of V is executing a biased Brownian motion:

${d\over dt} \log(V) = (-\gamma + i \Omega) + i\eta$


$V(t) = \exp( - (\gamma + i \Omega)t + iB(t))$

Where B is a standard Brownian motion. The V solution goes to zero for sure, even without the i multiplying the Brownian term, because B only goes as square root of t (assuming positive nonzero damping), and the total distance travelled is given by a Laplace transform of the exponential of a Brownian motion, and I didn't bother to evaluate its statistics because this is clearly not physics.

  • The x and y $\eta$ noise can't be the same in physics. I didn't realize they were the same in the problem as stated, because this is so strange.
  • In physical Brownian diffusion, in the limit of long times, V acquires a distribution with a scale, the Maxwell Boltzmann distribution, and in this case V can't acquire dimension because the equation is scale invariant (the OP says "dimensionally consistent"). There is a thermal velocity scale of $V_T=\sqrt{2mkT}$ in real Brownian motion which allows the equation to be dimensionally consistent without V multiplying the noise, but with $V_T$. It is this thermal velocity that sets the scale for the average velocity at times long compared to the velocity relaxation time.

The Smoluchowski approximation requires that V have a steady state distribution and a relaxation time to that steady state distribution, which doesn't happen in this problem. Here the velocity goes away. That's clearly a property of the scaling invariance of V, and the overall decay of V, so it won't be fixed even if the x and y noise are different. This is not thermal physics, and I can't see any way to do a Smoluchowski approximation, so how to do the approximation is moot.

It is possible that the book had a thermal scale constant V multiplying the noise, and there is some error in transcription of the problem, but I don't want to start guessing.

Later: Correct stuff

Thanks for linking to the book, it made everything clear. The key words are "energy conserving process". This is designed to model a particle moving through a constant magnetic field with an additional random magnetic field on top, so that it goes around with a randomly fluctuating curvature, but the particle never changes speed.

Disregard the previous stuff, it is totally wrong, because it is using a different convention for the time derivative of stochastic quantities than is used throughout the book (I should have seen this from the form of the problem, but I didn't). In deriving the solution I used the fact that

${1\over V} {dV\over dt} = {d\over dt} \log(V)$

which is only true if the time derivative is Stratonovich (centered difference), which is a common convention in physics. The book is using an Ito convention for time derivatives (forward-difference), which is the most common convention in mathematics and in finance. The Ito time derivative does not obey the uncorrected chain rule, but the correction is easy to work out:

${1\over V(t+\epsilon)} {dV\over dt} -{ 1\over V(t)} {dV\over dt} = -{1\over V^2} {(V(t+\epsilon) - V(t))^2\over \epsilon} $

and the last term, the infinitesimal square fluctuation in V at time t, is entirely determined by the square of the coefficient of the noise term. In quantum mechanics language,

$[{1\over V}, {dV\over dt} ] = (i\sqrt{2\gamma})^2 = -2\gamma$

In order to go from Ito derivative, which is one order of operators, to Stratonovich derivative which is the average of the two orders (like the Jordan product of operators in quantum mechanics), you have to add half the commutator, and this gives

${d\over dt} \log(V) - \gamma = -\gamma + i\Omega + i\sqrt{2\gamma}\eta$

Which exactly cancels the damping term, so that the result is completely physical--- the velocity magnitude doesn't decay at all. This is not a surprise. The coefficient which I confused for a physical damping is not a damping at all, but just the amount of magnitude change of the velocity vector an infinitesimal time into the future caused by the random direction-changing fluctuations.

$V = V_0 \exp(i\Omega t + i\sqrt{2\gamma}B(t))$

This means that the angle of the velocity V is executing a Brownian motion with drift, and keeps the same magnitude at all times. The equation at long times does become a Brownian motion in x, and the Smoluchowski approximation is valid for long times.

The Smoluchowski limit

The central difficulty you have is with the Smoluchowski limit, so I'll give a short explanation. In ordinary Brownian motion, the equation of motion is:

$m{d^2x\over dt^2} + \gamma{dx\over dt} + \sqrt{2D}\eta = 0$

(Ito and Stratonovich are the same, there are no products of non-commuting quantities). If you average this equation over long time window, the velocity averages to the total distance over the total time, while the second derivative averages to zero (because the velocity is varying in a bounded thermal range). This means that the first derivative average is equal to the noise thermal average, which means that the long-time averaged displacement of the solution to the above second order stochastic equation is equal to the long-time averaged solution of the first order equation below, which drops the second derivative term:

$\gamma {dx\over dt} + \sqrt{2D}\eta =0$

Or, more succinctly at the cost of butchering history, in Brownian motion, Newton equals Aristotle. So that $D$ is the diffusion constant of the long-time Brownian motion (D is also determined from the condition that the v diffusion process has as a steady state the Maxwell Boltzmann distribution, and this gives the Einstein relation). Dropping the second derivative term in the equation is equivalent to the formal rules that the book gives. There is no more content in those formal rules than the above averaging procedure, which is better because its both clearer and more rigorous.

The actual problem

In this problem, the equation in question is (in second order Ito form--- all derivatives are forward differences)

${d^2 X\over dt^2} - \gamma {dX\over dt} + i\Omega{dX\over dt} - i\sqrt{2\gamma}{dX\over dt} \eta = 0$

The Smoluchowski approximation averages this equation over a long interval, which gets rid of the second derivative term (for the same reason as before, bounded velocity fluctuations), and produces an equation which is first order:

$(-\gamma + i\Omega)\Delta X = i\sqrt{2\gamma}\int_0^T {dX\over dt}\eta$

But unlike standard thermal drift, this is no simplification, because one needs to know the average of the product of the velocity and the noise in order to get a closed equation. This is the difficulty you were having.

There is a minor annoyance in this way of looking at things: it seems like there are too many equations, in that the real and imaginary parts of the equation give two separate constraints, but this isn't true, because the right hand side is almost a perfect integral. By putting the explicit solution for the velocity,

$\int_0^T i\sqrt{2\alpha} {dX\over dt}\eta(t) dt = \int_0^T \sqrt{2\alpha}V_0\eta e^{i\Omega t + i\sqrt{2\alpha} B(t)} dt$

And considering that $\eta = {dB\over dt}$ by definition, this would be a perfect integral up to adding Omega, in the stratonovich convention, where the chain rule works. But this is Ito, so you do the half-commutator trick etc, etc, I'll leave out the details, but the end result gives just one consistent equation for $\Delta X$

$\Delta X = V_0 \int_0^T \exp(i\Omega t + B(t)) dt$

which tells us nothing new, because this is just the integral of the velocity from the explicit solution. So there is no way to evaluate the Smoluchowski limit without knowing something about the integral of cosines and sines of a Brownian motion. This is the central problem that the book is posing.

The integral of cosines and sines of a Brownian motion

To get the Smolochowski limit, all you need evaluate the average square distance traveled after time t:

$\langle |\Delta X|^2 \rangle = |V_0|^2\langle \int_0^T \int_0^T e^{i\Omega (t - t') + i (B(t)-B(t'))} dt dt' \rangle = \int_0^T\int_0^T e^{i\Omega(t-t')} \langle e^{i(B(t)-B(t'))}\rangle$

So the irreducible quantity to solve this problem is:

$\langle e^{i(B(t) - B(t'))} \rangle = G(t-t')$

$G(t)$ is, by translation invariance, the expected value of

$G(t) = \langle e^{iB(t)} \rangle $

This can be evaluated by Feynman diagrams: expand the exponential in powers, and note that B(t) is gaussian with width t, so all its moments are known. The odd moments are zero, and the even moments are given by products of odd numbers (pairing combinations):

$\langle B(t)^{2n} \rangle = 1 \cdot 3 \cdot 5 ...\cdot (2n-1) t^n$

So the power series for the exponential only has odd terms

$G(t) = \sum_{k=0}^\infty {(-1)^n \langle B(t)^{2n}\rangle\over (2n)!} = e^{-{t\over 2}}$

And this means by symmetry that

$G(t) = e^{-{1\over 2} |t-t'|}$

This can be found by path integrals too, it is a quadratic path integral with a linear source, and the result is the exponential of 1/2 the Green's function of a 1d laplacian between the two sources, and this Green's function is the absolute value function.

The distance traveled after time T is the double integral

$|\Delta X|^2 = |V_0|^2\int_0^T \int_0^T G(t-t') e^{i\Omega(t-t')}$

which by thinking about it in 45-degree rotated coordinates, is, up to the unimportant boundaries of the rectangle, equal to:

$|\Delta X|^2 = T|V_0|^2 \tilde{G}(\Omega)$

So that the diffusion constant is the Fourier transform of $\exp(-|t|)$, which is ${1\over 1+\omega^2}$, fixed up to account for the factor of 2. This gives the average square distance, and the diffusion constant is the coefficient of T in this expression:

$\tilde{G}(\Omega) = {1\over 4 + \Omega^2}$

You can restore the $\gamma$ into the formula by dimensional analysis, I'll do so later if necessary.

  • $\begingroup$ Thanks for the detail. What exactly do you mean when you say "this is clearly not physics"? The original problem can be found as Problem 10.2 here:… (No idea about the provenance of that file. I worked from a paper edition.) $\endgroup$ – Dan Piponi Aug 23 '11 at 13:04
  • $\begingroup$ Once I saw the book, I realized the problem was a charged particle in 2d with a perpendicular magnetic field which has a mean value but also varies randomly in space and time at short scales. I sketched the solution to the problem, but it requires an essential datum: the expected value of the cosine of a Brownian motion, which you can't get by formal tricks. $\endgroup$ – Ron Maimon Aug 24 '11 at 5:20
  • $\begingroup$ Fantastic! I'm actually not going to read too closely yet. I'll try to figure out the cosine of Brownian motion by myself. But I wouldn't have thought to do that without your hints. The weird thing is that every single problem in this book has been trivial (modulo mechanical calculation) apart from this one. $\endgroup$ – Dan Piponi Aug 24 '11 at 15:15

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